## Related questions with answers

Annual salary plus bonus data for chief executive officers are presented in the $BusinessWeek$ Annual Pay Survey. A preliminary sample showed that the standard deviation is $\$675$ with data provided in thousands of dollars. How many chief executive officers should be in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of $\$100,000$? (Note: The desired margin of error would be $E=100$ if the data are in thousands of dollars.) Use $95\%$ confidence.

Solutions

VerifiedWe have given:

- standard deviation of $\sigma=675$,
- desire margin of error of $E=100$.

Our task will be to determine the sample size which provides the desired margin of error at the $(1-\alpha)=95\%$ confidence level.

Given:

$c=95\%$

$E=100$

$\sigma=675$

Formula sample size:

$n=\left( \dfrac{z_{\alpha/2}\sigma}{E}\right)^2$

For confidence level $1-\alpha=0.95$, determine $z_{\alpha/2}=z_{0.025}$ using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

$z_{\alpha/2}=1.96$

The sample size is then (round up!):

$n=\left( \dfrac{z_{\alpha/2}\sigma}{E}\right)^2=\left( \dfrac{1.96\times 675}{100}\right)^2\approx 176$

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